Convolution Responses for Unequal Transit Rates

Why Unequal Transit Rates Matter

A cascade can delay a signal even when each stage responds by a simple first-order process. When the stages have different rate constants, however, the resulting output is not described by the identical-stage formula alone: each stage contributes a different exponential timescale. Convolution provides a direct way to combine those contributions and derive the response of the full chain.

The aim here is to derive responses for two or three first-order stages with distinct rates, then use them to calculate outputs for idealized inputs. These results apply to a linear, time-invariant chain. They do not by themselves specify how a drug generates the cascade input; that question depends on the upstream model.

Definition: A first-order transit stage with rate constant \(k_i\) has a normalized impulse-response kernel \(g_i(t)=k_i e^{-k_i t}\) for \(t\geq 0\), and \(g_i(t)=0\) for \(t<0\). Here \(k_i>0\) has units of inverse time. For a chain of \(m\) stages, the overall impulse response is the convolution \(h_m=g_1*g_2*\cdots*g_m\). The output for input \(u(t)\) is \(y(t)=\int_0^t h_m(t-\tau)u(\tau)\,d\tau\), assuming zero initial conditions.

This formulation assumes that each stage is linear, has a constant positive rate, and receives the preceding stage's output. The cascade has no direct bypass from input to output, and all stages start at zero in the modeled response (or are expressed as deviations from baseline). The input and output use the same signal units, while \(h_m\) has units of inverse time. An impulse input \(A\delta(t)\) has area \(A\), where \(A\) has units of signal multiplied by time; its output is \(A h_m(t)\).

Deriving the Unequal-Rate Responses

For two stages, convolution gives

h_2(t)=\int_0^t k_1 e^{-k_1\tau} k_2 e^{-k_2(t-\tau)}\,d\tau,\qquad t\geq 0.

When \(k_1\ne k_2\), factor out the terms that do not depend on \(\tau\), integrate the remaining exponential, and simplify:

\begin{aligned} h_2(t) &= k_1k_2e^{-k_2t}\int_0^t e^{(k_2-k_1)\tau}\,d\tau\\ &= k_1k_2e^{-k_2t}\frac{e^{(k_2-k_1)t}-1}{k_2-k_1}\\ &= \frac{k_1k_2}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right). \end{aligned}

The expression is unchanged if the stage labels are swapped, even though its displayed form changes. Its convolution definition also shows that \(h_2(t)\) is nonnegative; this remains true if the rates are entered in the opposite order.

For three distinct rates, the Laplace transform of each kernel is \(k_i/(s+k_i)\), so the transform of the cascade is the product. Partial fractions then give a compact exponential sum:

\begin{aligned} H_3(s)&=\frac{k_1k_2k_3}{(s+k_1)(s+k_2)(s+k_3)},\\ h_3(t)&=\sum_{i=1}^{3} \frac{k_1k_2k_3}{\displaystyle\prod_{j\ne i}(k_j-k_i)} e^{-k_i t},\qquad t\geq 0. \end{aligned}

For example, the coefficient multiplying \(e^{-k_i t}\) is found by multiplying \(H_3(s)\) by \(s+k_i\) and evaluating at \(s=-k_i\). The denominator then contains \(k_j-k_i\) for each other stage, which fixes the signs as well as the coefficient values. Although the three terms can have alternating signs, their sum is nonnegative because it is the convolution of nonnegative kernels.

The same Laplace-transform argument gives, for \(m\) distinct positive rates, the sum with coefficient \( \prod_{j=1}^m k_j \big/ \prod_{j\ne i}(k_j-k_i)\) on \(e^{-k_i t}\). The two- and three-stage forms are usually easier to interpret directly. If rates coincide, these distinct-rate formulas have zero denominators; the coincident-rate limit produces exponential terms multiplied by powers of time, as in the material on identical transit stages.

Worked Example: Two Stages With Unequal Rates

Suppose a hypothetical two-stage chain has \(k_1=0.500\ \mathrm{h}^{-1}\) and \(k_2=1.50\ \mathrm{h}^{-1}\). Find its impulse response at \(t=1.00\ \mathrm{h}\), and locate its peak.

The rates are positive and distinct, so the two-stage formula applies. Substitution gives

\begin{aligned} h_2(1) &=\frac{(0.500)(1.50)}{1.50-0.500} \left(e^{-(0.500)(1)}-e^{-(1.50)(1)}\right)\\ &=0.750\left(0.606531-0.223130\right)\\ &\approx 0.2876\ \mathrm{h}^{-1}. \end{aligned}

To find the peak, differentiate the response and set the derivative equal to zero:

h_2'(t)=\frac{k_1k_2}{k_2-k_1} \left(-k_1e^{-k_1t}+k_2e^{-k_2t}\right)=0.

For \(k_2>k_1\), rearranging gives \(e^{(k_2-k_1)t}=k_2/k_1\), and hence \(t_{\mathrm{peak}}=\ln(k_2/k_1)/(k_2-k_1)=\ln(3)/1.00\approx1.099\ \mathrm{h}\). The response is still rising at one hour, consistent with this peak time. The rate difference affects when the maximum occurs; neither stage acts as an instantaneous delay.

Worked Example: Three Distinct Rates

Consider a hypothetical three-stage chain with \(k_1=1.00\ \mathrm{h}^{-1}\), \(k_2=2.00\ \mathrm{h}^{-1}\), and \(k_3=4.00\ \mathrm{h}^{-1}\). Derive the response and evaluate it at one hour.

The rates are all positive and distinct. The three coefficients are \(8/[(2-1)(4-1)]=8/3\), \(8/[(1-2)(4-2)]=-4\), and \(8/[(1-4)(2-4)]=4/3\), in the corresponding exponential terms. Thus

h_3(t)=\frac{8}{3}e^{-t}-4e^{-2t}+\frac{4}{3}e^{-4t}\quad \mathrm{h}^{-1},

where \(t\) is in hours. At one hour,

\begin{aligned} h_3(1) &=\frac{8}{3}(0.367879)-4(0.135335)+\frac{4}{3}(0.018316)\\ &\approx 0.981011-0.541341+0.024421\\ &\approx 0.4641\ \mathrm{h}^{-1}. \end{aligned}

The terms have different signs, but the final response is positive, as required by its convolution definition. The slowest rate, \(1.00\ \mathrm{h}^{-1}\), governs the long-time tail: after the faster exponentials have largely decayed, the leading behavior is approximately \((8/3)e^{-t}\ \mathrm{h}^{-1}\). This late-time approximation is not an accurate description near the start of the response, when all three terms matter.

Using Convolution for a Finite Input

An impulse response is useful beyond an idealized impulse. For an input that is constant at level \(U\) from time zero to time \(T\), and zero afterward, the output during the input interval is \(U\int_0^t h_m(v)\,dv\) for \(0\leq t\leq T\). This follows by substituting the constant input into the convolution integral and changing the integration variable. The integral is the step response: it accumulates the impulse response up to the time of interest.

Worked Example: Output During a Constant Three-Stage Input

Use the three rates \(1.00\), \(2.00\), and \(4.00\ \mathrm{h}^{-1}\) above. Suppose a hypothetical input is \(U=2.00\) signal units from time zero through \(T=2.00\ \mathrm{h}\). Find the output at \(t=2.00\ \mathrm{h}\), just before the input ends.

The input is constant over the entire integration interval, so

\begin{aligned} y(2) &=2.00\int_0^2 h_3(v)\,dv\\ &=2.00\left[ \frac{8}{3}(1-e^{-2})-2(1-e^{-4})+\frac{1}{3}(1-e^{-8}) \right]\ \mathrm{signal}. \end{aligned}

Using \(e^{-2}=0.1353353\), \(e^{-4}=0.0183156\), and \(e^{-8}=0.0003355\), the accumulated response is

\begin{aligned} \int_0^2h_3(v)\,dv &=1-\frac{8}{3}(0.1353353)+2(0.0183156)-\frac{1}{3}(0.0003355)\\ &\approx 0.6756254,\\ y(2)&=(2.00)(0.6756254)\approx1.351\ \mathrm{signal}. \end{aligned}

The result has signal units because the integral of an impulse response in inverse hours over hours is dimensionless. It is the output at the end of the input interval, not the eventual output after the input has stopped and the cascade has continued to respond.

Checks, Diagnostics, and Comparisons

Several checks help detect algebra or implementation errors. For positive rates, each stage kernel integrates to one, so the area under the overall impulse response is also one. This follows because the integral of a convolution is the product of the integrals of its factors. In the exponential-sum formulas, the same check is \(H_m(0)=1\), or equivalently the sum of each exponential coefficient divided by its rate is one. The response also has the units of inverse time and must be nonnegative for every \(t\geq0\).

The shape reflects the relative timescales. A small rate produces a slowly decaying component and a longer tail; a large rate decays quickly. The full response is not simply the slowest stage's exponential, because the chain passes through every stage. Nor does a distinct-rate chain have an output that peaks at the input time: transit through multiple stages spreads the response over time. For a three-stage chain, alternating signs in the closed form are normal, but a negative computed value signals a likely sign error, inadequate numerical precision, or an implementation problem.

Approach	Useful when	Main caution
Closed-form exponential sum	Rates are distinct and the analytic shape or an exact convolution calculation is needed	Nearly equal rates can cause numerical cancellation
Direct convolution integral	The input is specified and a response can be evaluated by integration	Requires accurate integration over the input history
Numerical integration of the stage equations	Many stages or more complicated inputs make a closed form inconvenient	Results depend on appropriate solver tolerances and time resolution

When two rates are very close, the coefficients in a partial-fraction expression can become large and nearly cancel, even though the actual response remains finite. In that case, evaluate the convolution directly, use a stable limiting expression, or solve the stage equations numerically. Agreement between the closed form and a numerical solution at a few test times is a useful diagnostic. A model with time-varying rates, nonlinear transfer, nonzero initial deviations, or feedback does not satisfy the assumptions used for these formulas without further modification.

Common Pitfall

A frequent error is to read the exponential-sum terms as separate physical responses and interpret a negative coefficient as a negative stage output. The individual terms are components of an algebraic representation, not independent outputs. Their sum is the cascade impulse response, which is nonnegative because it is a convolution of nonnegative kernels. Check the complete sum, its initial behavior, units, and total area; do not interpret a partial-fraction coefficient in isolation.

Key takeaway: For distinct positive transit rates, convolution produces a sum of exponentials with coefficients determined by rate differences. Rate heterogeneity changes the peak and tail, while the response retains nonnegativity and unit area under the stated assumptions.

Check Your Understanding

Answer each question, then compare with the solutions below.

For two stages with \(k_1=0.4\ \mathrm{h}^{-1}\) and \(k_2=1.2\ \mathrm{h}^{-1}\), write the impulse-response formula. Is it nonnegative for \(t\geq0\)?
For the three-stage rates \(1\), \(2\), and \(4\ \mathrm{h}^{-1}\), which rate controls the late-time tail, and why?
Why must the integrated impulse response equal one for a chain of normalized stages?
What problem can arise when two distinct rates are numerically very close, and name one suitable response.

Solutions. (1) \(h_2(t)=0.6(e^{-0.4t}-e^{-1.2t})\ \mathrm{h}^{-1}\). For \(t\geq0\), \(e^{-0.4t}\geq e^{-1.2t}\), so the response is nonnegative. (2) The \(1\ \mathrm{h}^{-1}\) rate controls the tail because its exponential decays most slowly; the faster terms eventually become negligible. (3) Each kernel integrates to one, and the integral of a convolution equals the product of the integrals, so the chain response also integrates to one. (4) Large, nearly cancelling coefficients can cause numerical loss of precision. Direct convolution or numerical integration of the stage equations can avoid unstable evaluation of the exponential sum.